Past exam of the topics in applied mathematics course of the University of Oxford / 2010 / A / 1 / b

A steady population $N_{\ast }$ is a root $h(N_{\ast })=0$. To linearize one needs $h$ differentiable near that root, an extra local regularity assumption beyond the source's continuity. Setting $N=N_{\ast }+\eta$ gives $\dot{\eta }=h^{\prime }(N_{\ast })\eta +o(\eta )$, hence $\eta (t)=\eta (0)e^{h^{\prime }(N_{\ast })t}$ at linear order.

Thus $h^{\prime }(N_{\ast })<0$ gives linear asymptotic stability and $h^{\prime }(N_{\ast })>0$ gives instability. If $h^{\prime }(N_{\ast })=0$, linearization is inconclusive. In a one-dimensional phase line, the more general nonlinear test is whether the vector field points towards the root from both sides. At the physical boundary $N=0$, only the right-hand side matters. Continuity alone does not guarantee uniqueness of trajectories, so a standard well-defined flow also presumes a local uniqueness condition, such as local Lipschitz continuity.

The four nonnegative equilibria are $0,a_1,a_2,a_3$. Since all factors are positive below $a_1$, the sign of $h(N)$ on consecutive intervals isThe phase line therefore points away from zero, towards $a_1$, away from $a_2$, and towards $a_3$. Equivalently $h^{\prime }(0)=a_1{a}_2{a}_3>0$, and at each simple positive root $a_i$, $h^{\prime }(a_i)=-a_i{\prod }_{j\ne i}(a_j-a_i)$. Their signs alternate negative, positive, negative.

Hence extinction and $a_2$ are unstable, while $a_1$ and $a_3$ are asymptotically stable. Initial positive populations below $a_2$ tend to $a_1$; those above $a_2$ tend to $a_3$; an initial population exactly at $a_2$ remains there. This is bistability with a threshold separating the two basins.

Put $P(N)=(a_1-N)(a_2-N)(a_3-N)$. The harvested vector field is $N[P(N)-E]$. Define the extinction threshold $E_T=P(0)=a_1{a}_2{a}_3$, and let $N_M\in (a_2,a_3)$ be the positive local maximum of $P$, with $E_M=P(N_M)>0$. Writing $s_1=a_1+a_2+a_3$, $s_2=a_1{a}_2+a_1{a}_3+a_2{a}_3$ givesPositive equilibria are intersections of $P(N)$ with the horizontal line $E$. A lower stable branch in $(0,a_1)$ exists exactly for $0<E<E_T$. A pair in $(a_2,a_3)$ exists exactly for $0<E<E_M$: its smaller member has $P^{\prime }>0$ and is unstable; its larger member has $P^{\prime }<0$ and is stable. At a positive root the linear eigenvalue is $N{P}^{\prime }(N)$, which proves these claims. Extinction has eigenvalue $E_T-E$, becoming stable for $E>E_T$.

The harvested bistable population thresholds determine the behaviour. If $E_M<E_T$, the upper stable population and its unstable separator first disappear in a saddle-node bifurcation. Populations then tend to the remaining lower stable state until that branch reaches extinction at $E_T$; beyond it extinction attracts every positive initial state. If $E_T<E_M$, the lower stable branch reaches extinction first. For $E_T<E<E_M$, extinction and the upper positive state are both stable, separated by the middle unstable root. Increasing effort through $E_M$ then destroys the remaining positive attractor, causing a sudden collapse to extinction. If $E_T=E_M$, both events occur at the same effort at different populations.

At $E=E_T$, near zero the leading term is $P^{\prime }(0)N^2<0$, so extinction attracts from the physical right side algebraically, despite its zero linear eigenvalue. At $E=E_M$, the double positive root is semistable: locally $\dot{N}\simeq -\frac{1}{2}{N}_M|P^{\prime \prime }(N_M)|(N-N_M{)}^2$, attracting from above and repelling from below. These critical cases matter for exact initial data. The figure shows examples of both strict threshold orderings.